Question

The normal at $(1,1)$ to the circle $x^2+y^2-4x+6y=0$ is

• A. $4x+3y=7$
• B. $4x+y=5$
• C. $x+y=2$
• D. $4x+y=5$

Answer 1

The correct option is B $4x+y=5$

The equation of circle is given as,

$x^2+y^2-4x+6y=0$

$\therefore (x^2-4x)+(y^2+6y)=0$

$\therefore (x^2-4x+4)-4+(y^2+6y+9)-9=0$

$\therefore {(x-2)}^2+{(y+3)}^2=13$

$\therefore {(x-2)}^2+{[y-(-3)]}^2={\left(\sqrt{13}\right)}^2$

Compare this equation with standard equation of circle i.e. $\therefore {(x-h)}^2+{[y-k]}^2=r^2$, we get

$h=2$, $k=-3$ and $r=\sqrt{13}$

Thus, center of circle is, $C(2,-3)$ and radius is $r=\sqrt{13}$

Now, Let tangent to the circle is drawn through point $A(1,1)$

as given in problem.

Let $m_1$ = slope of radius AC

Let $m_2$ = Slope of tangent

$\therefore m_1\times m_2=-1$ Equation (1)

By two point form, slope of radius AC is calculated as,

$m_1=\frac{-3-1}{2-1}$

$\therefore m_1=\frac{-4}{1}=-4$

From equation (1),

$-4\times m_2=-1$

$\therefore m_2=\frac{1}{4}$

Now, Let $m_3$ = slope of normal through point $A(1,1)$

As tangent and normal passing through $(1,1)$ are perpendicular to each other, we can write,

$m_2\times m_3=-1$

$\therefore \frac{1}{4}\times m_3=-1$

$\therefore m_3=-4$

Thus, equation of normal passing through $(1,1)$ is given by two point form as,

$y-y_1=m_3(x-x_1)$

$\therefore y-1=-4(x-1)$

$\therefore y-1=-4x+4$

$\therefore 4x+y=5$

Thus, answer is option (B)